Practice MCQ Questions and Answer on Volume and Surface Area

1.

Water is poured into an empty cylindrical tank at a constant rate for 5 minutes. After the water has been poured into the tank. the depth of the water is 7 feet. The radius of the tank is 100 feet. Which of the following is the best approximation for the rate at which the water was poured into the tank ?

• (A) 140 cubic feet/sec
• (B) 440 cubic feet/sec
• (C) 700 cubic feet/sec
• (D) 2200 cubic feet/sec

Show Answer

 Volume of water flown into the tank in 5 min : $$\eqalign{ & = \left( {\frac{{22}}{7} \times 100 \times 100 \times 7} \right){\text{cu}}{\text{.feet}} \cr & = 220000\,{\text{cu}}{\text{.feet}} \cr} $$ ∴ Rate of flow of water : $$\eqalign{ & = \left( {\frac{{220000}}{{5 \times 60}}} \right){\text{cu}}{\text{.feet/sec}} \cr & = 733.3 \approx 700\,{\text{cu}}{\text{.feet/sec}} \cr} $$

2.

If three equal cubes are placed adjacently in a row, then the ratio of the total surface area of the new cuboid to the sum of the surface areas of the three cubes will be ?

• (A) 1 : 3
• (B) 2 : 3
• (C) 5 : 9
• (D) 7 : 9

Show Answer

 Let the length of each edge of each cube be a Then, the cuboid formed by placing 3 cubes adjacently has the dimensions 3a , a and a Surface area of the cuboid : $$\eqalign{ & = 2\left[ {3a \times a + a \times a + 3a \times a} \right] \cr & = 2\left[ {3{a^2} + {a^2} + 3{a^2}} \right] \cr & = 14{a^2} \cr} $$ Sum of surface area of 3 cubes : $$\eqalign{ & = \left( {3 \times 6{a^2}} \right) \cr & = 18{a^2} \cr} $$ ∴ Required ratio : $$\eqalign{ & = 14{a^2}:18{a^2} \cr & = 7:9 \cr} $$

3.

The external and internal diameters of a hemispherical bowl are 10 cm and 8 cm respectively. What is the total surface area of the bowl ?

• (A) 257 cm2
• (B) 286 cm2
• (C) 292 cm2
• (D) 302 cm2

Show Answer

 Internal radius, r = 4 cm External radius, R = 5 cm Total surface area : $$\eqalign{ & = 2\pi {R^2} + 2\pi {r^2} + \pi \left( {{R^2} - {r^2}} \right) \cr & = 3\pi {R^2} + \pi {r^2} \cr & = \left[ {\pi \left( {3 \times 25 + 16} \right)} \right]{\text{ c}}{{\text{m}}^2} \cr & = \left( {\frac{{22}}{7} \times 91} \right){\text{c}}{{\text{m}}^2} \cr & = 286{\text{ c}}{{\text{m}}^2} \cr} $$

4.

A well with inner diameter 8 m is dug 14 m deep. Earth take out of its has been evenly spread all around it to a width of 3 m form an embankment. The height of the embankment will be :

• (A) $$\text{4}\frac{26}{33}\text{m}$$
• (B) $$\text{5}\frac{26}{33}\text{m}$$
• (C) $$\text{6}\frac{26}{33}\text{m}$$
• (D) $$\text{7}\frac{26}{33}\text{m}$$

Show Answer

 Volume of earth dug out : $$\eqalign{ & = \left( {\frac{{22}}{7} \times 4 \times 4 \times 4} \right){{\text{m}}^3} \cr & = 704\,{{\text{m}}^3} \cr} $$ Area of embankment : $$\eqalign{ & = \frac{{22}}{7} \times \left( {{7^2} - {4^2}} \right) \cr & = \left( {\frac{{22}}{7} \times 11 \times 3} \right){{\text{m}}^3} \cr & = \frac{{726}}{7}{{\text{m}}^3} \cr} $$ Height of embankment : $$\eqalign{ & = \left( {\frac{{{\text{Volume}}}}{{{\text{Area}}}}} \right) \cr & = \left( {\frac{{704 \times 7}}{{726}}} \right){\text{m}} \cr & = \frac{{224}}{{33}}{\text{m}} \cr & = {\text{ 6}}\frac{{26}}{{33}}{\text{m}} \cr} $$

5.

A circular well with a diameter of 2 metres, is dug to a depth of 14 metres. What is the volume of the earth dug out ?

• (A) 32 m3
• (B) 36 m3
• (C) 40 m3
• (D) 44 m3

Show Answer

 Volume : $$\eqalign{ & = \pi {r^2}h \cr & = \left( {\frac{{22}}{7} \times 1 \times 1 \times 14} \right){m^3} \cr & = 44\,{m^3} \cr} $$

6.

If the heights of two cones are in the ratio 7 : 3 and their diameters are in the ratio 6 : 7, what is the ratio of their volumes ?

• (A) 3 : 7
• (B) 4 : 7
• (C) 5 : 7
• (D) 12 : 7

Show Answer

 Let the heights of two cones be 7x and 3x and their radii be 6y and 7y respectively Then, Ratio of volume : $$\eqalign{ & = \frac{{\frac{1}{3}\pi \times {{\left( {6y} \right)}^2} \times 7x}}{{\frac{1}{3}\pi \times {{\left( {7y} \right)}^2} \times 3x}} \cr & = \frac{{36 \times 7}}{{49 \times 3}} \cr & = \frac{{12}}{7}\,Or\,12:7 \cr} $$

7.

A closed box made of wood of uniform thickness has length, breadth and height 12 cm, 10 cm and 8 cm respectively. If the thickness of the wood is 1 cm, the inner surface area is :

• (A) 264 cm2
• (B) 376 cm2
• (C) 456 cm2
• (D) 696 cm2

Show Answer

 Internal length = (12 - 2) cm = 10 cm Internal breadth = (10 - 2) cm = 8 cm Internal height = (8 - 2) cm = 6 cm Inner surface area : = 2 [10 × 8 + 8 × 6 + 10 × 6] cm2 = (2 × 188) cm2 = 376 cm2

8.

A hall is 15 m long and 12 m broad. If the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls, the volume of the hall is:

• (A) 720 m3
• (B) 900 m3
• (C) 1200 m3
• (D) 1800 m3

Show Answer

 $$\eqalign{ & 2\left( {15 + 12} \right) \times h = 2\left( {15 \times 12} \right) \cr & \Rightarrow h = \frac{{180}}{{27}}m = \frac{{20}}{3}m \cr & \therefore {\text{Volume}} = \left( {15 \times 12 \times \frac{{20}}{3}} \right){m^3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1200\,{m^3} \cr} $$

9.

Water flows out through a circular pipe whose internal diameter is 2 cm, at the rate of 6 metres per second into a cylindrical tank, the radius of whose base is 60 cm. By how much will the level of water rise in 30 minutes ?

• (A) 2 m
• (B) 3 m
• (C) 4 m
• (D) 5 m

Show Answer

 Volume of water flown through the pipe in 30 min : $$\eqalign{ & = \left[ {\left( {\pi \times 0.01 \times 0.01 \times 6} \right) \times 30 \times 60} \right]{m^3} \cr & = \left( {1.08\pi } \right){m^3} \cr} $$ Let the rise in level of water be h metres Then, $$\eqalign{ & \pi \times 0.6 \times 0.6 \times h = 1.08\pi \cr & \Rightarrow h = \left( {\frac{{1.08}}{{0.6 \times 0.6}}} \right) \cr & \Rightarrow h = 3\,m \cr} $$

10.

A solid metallic right circular cylinder of base diameter 16 m and height 2 cm is melted and recast into a right circular cone of height three times that of the cylinder. Find the curved surface area of the cone. [Use $$\pi$$ = 3.14]

• (A) 196.8 cm2
• (B) 228.4 cm2
• (C) 251.2 cm2
• (D) None of these

Show Answer

 Let the radius of the cone be r cm Then, $$\eqalign{ & \pi \times {\left( 8 \right)^2} \times 2 = \frac{1}{3} \times \pi \times {r^2} \times 6 \cr & \Rightarrow r = 8 \cr} $$ Slant height, $$\eqalign{ & l = \sqrt {{r^2} + {h^2}} \cr & \,\,\, = \sqrt {{8^2} + {6^2}} \cr & \,\,\, = \sqrt {100} \cr & \,\,\, = 10\,cm \cr} $$ Curved surface area of cone : $$\eqalign{ & = \pi rl \cr & = \left( {3.14 \times 8 \times 10} \right){\text{ c}}{{\text{m}}^2} \cr & = 251.2{\text{ c}}{{\text{m}}^2} \cr} $$